# Calculating energy eigenvalues when potential is given

So our teacher claimed that if we have a Potential of the form $V(x)= x^\nu$ then the Energy is of the form $E={2\nu \over \nu+2}$ Can anyone break up the math for this problem?

• Do you mean the ground state energy? If $\nu$ is an odd integer the energy is not bounded below. – Keith McClary Oct 8 '15 at 3:32
• Yes the bound state energy – Siddhartha Dam Oct 8 '15 at 4:31
• $x^\nu$ in imaginary if $x<0$ and $\nu$ is $\frac{1}{2}$. – Keith McClary Oct 8 '15 at 4:42
• V actually varies with x in,this manner.. So you can ignore the non trivial solutions – Siddhartha Dam Oct 8 '15 at 9:55

## 1 Answer

Assuming you consider only even powers of $x$ (for odd powers there are no bound states) the behaviour you describe comes from the Born-Sommerfeld condition. This is described in the Wikipedia article Old quantum theory, and there is a related question on this site.

Suppose $E$ is the energy of your system, and $\pm X$ are the turning points where $E = V(X)$, then for your system the Born-Sommerfeld condition states:

$$\int_{-X}^{+X} p \, dx = nh$$

where the momentum $p$ is given by:

$$p(x) = \sqrt{2m(E - V(x))}$$